wyznaczyć k:
\(\displaystyle{ {5 \choose k+1}^{2} = {5 \choose k} {6 \choose k+2}}\)
symbol newtona - wyznacz k
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JakimPL
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symbol newtona - wyznacz k
Wiadomo, że \(\displaystyle{ k \le 4}\). Po kolei będziemy upraszczać wyrażenia.
\(\displaystyle{ {5 \choose k+1}^{2} = {5 \choose k} {6 \choose k+2}}\)
\(\displaystyle{ \left(\frac{5!}{(k+1)!(4-k)!} \right)^{2} = \left(\frac{5!}{k!(5-k)!} \right)\left(\frac{6!}{(k+2)!(4-k)!} \right)}\)
\(\displaystyle{ \frac{5! \cdot 5! }{(k+1)!(k+1)!(4-k)!(4-k)!} = \frac{5! \cdot 6!}{k!(5-k)!(k+2)!(4-k)!}}\)
\(\displaystyle{ \frac{5! }{(k+1)!(k+1)!(4-k)!} = \frac{6!}{k!(5-k)!(k+2)!}}\)
\(\displaystyle{ \frac{5! }{(k+1)k!(k+1)!(4-k)!} = \frac{5! \cdot 6}{k!(5-k)!(k+2)!}}\)
\(\displaystyle{ \frac{1}{(k+1)(k+1)!(4-k)!} = \frac{6}{(5-k)!(k+2)!}}\)
\(\displaystyle{ \frac{1}{(k+1)(k+1)!(4-k)!} = \frac{6}{(5-k)!(k+1)!(k+2)}}\)
\(\displaystyle{ \frac{1}{(k+1)(4-k)!} = \frac{6}{(5-k)!(k+2)}}\)
\(\displaystyle{ \frac{k+2}{(4-k)!} = \frac{6(k+1)}{(4-k+1)!}}\)
\(\displaystyle{ \frac{k+2}{(4-k)!} = \frac{6(k+1)}{(4-k)!(4-k+1)}}\)
\(\displaystyle{ (k+2)(4-k+1) = 6(k+1)}\)
\(\displaystyle{ (k+2)(5-k) = 6(k+1)}\)
\(\displaystyle{ 5k - k^2 + 10 - 2k= 6k+6}\)
\(\displaystyle{ - k^2 + 3k + 10= 6k+6}\)
\(\displaystyle{ - k^2 - 3k + 4= 0}\)
\(\displaystyle{ k^2 + 3k - 4= 0}\)
\(\displaystyle{ (k+4)(k-1)= 0}\)
\(\displaystyle{ k = 1}\) spełnia założenia i taki jest wynik.
\(\displaystyle{ {5 \choose k+1}^{2} = {5 \choose k} {6 \choose k+2}}\)
\(\displaystyle{ \left(\frac{5!}{(k+1)!(4-k)!} \right)^{2} = \left(\frac{5!}{k!(5-k)!} \right)\left(\frac{6!}{(k+2)!(4-k)!} \right)}\)
\(\displaystyle{ \frac{5! \cdot 5! }{(k+1)!(k+1)!(4-k)!(4-k)!} = \frac{5! \cdot 6!}{k!(5-k)!(k+2)!(4-k)!}}\)
\(\displaystyle{ \frac{5! }{(k+1)!(k+1)!(4-k)!} = \frac{6!}{k!(5-k)!(k+2)!}}\)
\(\displaystyle{ \frac{5! }{(k+1)k!(k+1)!(4-k)!} = \frac{5! \cdot 6}{k!(5-k)!(k+2)!}}\)
\(\displaystyle{ \frac{1}{(k+1)(k+1)!(4-k)!} = \frac{6}{(5-k)!(k+2)!}}\)
\(\displaystyle{ \frac{1}{(k+1)(k+1)!(4-k)!} = \frac{6}{(5-k)!(k+1)!(k+2)}}\)
\(\displaystyle{ \frac{1}{(k+1)(4-k)!} = \frac{6}{(5-k)!(k+2)}}\)
\(\displaystyle{ \frac{k+2}{(4-k)!} = \frac{6(k+1)}{(4-k+1)!}}\)
\(\displaystyle{ \frac{k+2}{(4-k)!} = \frac{6(k+1)}{(4-k)!(4-k+1)}}\)
\(\displaystyle{ (k+2)(4-k+1) = 6(k+1)}\)
\(\displaystyle{ (k+2)(5-k) = 6(k+1)}\)
\(\displaystyle{ 5k - k^2 + 10 - 2k= 6k+6}\)
\(\displaystyle{ - k^2 + 3k + 10= 6k+6}\)
\(\displaystyle{ - k^2 - 3k + 4= 0}\)
\(\displaystyle{ k^2 + 3k - 4= 0}\)
\(\displaystyle{ (k+4)(k-1)= 0}\)
\(\displaystyle{ k = 1}\) spełnia założenia i taki jest wynik.